Given a Möbius Transformation $w=f(z)$ find $f(-\bar{z})$

Question

Let $w=f(z)$ be a Möbius Transformation, and let $\gamma,\Gamma\subset\mathbb{C}$ be the following curves:

$$\gamma=\{z\in\mathbb{C}\mid \Re(z)=0\}\\ \Gamma=\{w\in\mathbb{C}\mid |w-w_0|=r\}$$

Where $w_0\in\mathbb{C}$ and $r\in\mathbb{R^+}$. Given that $f(\gamma)=\Gamma$, find the value of $h(z)=f(-\bar{z})$ using $w,w_0$ and $r$.

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I know that $f(z)$ is a Möbius Transformation, therefore:

$$f(z)=\frac{az+b}{cz+d}$$

For some $a,b,c,d\in\mathbb{C}$. Now because $f(\gamma)=\Gamma$, I receive:

$$f(0)=\frac bd\\f(\bar\infty)=\frac ac$$

Therefore $\frac ac, \frac bd\in\Gamma$. Choosing any other $z\in\gamma$ and trying to work out the equations led me to some ugly algebra which I couldn't manage to handle. I am sure there is an easier way to handle this but I don't know how.

Thanks

Answer 1

$z$ and $-\bar z$ are symmetric about the line $\gamma$. Such points are conjugate.

Mobius transformations happen to preserve conjugate points. Thus $h(z)$ is symmetric to $f(z)$ relative to the circle $\Gamma$. This means they are the images of each other under circle inversion.

So we get $h(z)=w_0+\dfrac{r^2( w-w_0)}{|w-w_0|^2}$.